- Flashback of IX real Number's
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- Euclid's Division Lemma
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- Proof of Euclid's Division Lemma
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- HCF (Highest common factor)
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- What is Prime Numbers
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- What is Composite Numbers
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- Fundamental Theorom of Arithmetic
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- HCF and LCM by prime factorization method
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- Irrational Numbers
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- How to prove the irrational numbers or Rational numbers

- Real number problem and Solutions
- |
- Real number Worksheet
- |
- Real number problems
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- Real number Important questions

In this page we have *NCERT solutions for Class 10 Maths Chapter 1:Real Numbers* for
EXERCISE 1.1 . Hope you like them and do not forget to like , social_share
and comment at the end of the page.

(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

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We know two things from the chapter

__Euclid’s Division Lemma__

For a and b any two positive integer, we can always find unique integer q and r such that

a=bq + r , 0 ≤ r ≤ b

If r =0, then b is divisor of a.

__HCF (Highest common factor)__

HCF of two positive integers can be find using the Euclid’s Division Lemma algorithm

We know that for any two integers a,b. we can write following expression

a=bq + r , 0 ≤ r ≤ b

If r=0 ,then

HCF( a,b) =b

If r≠0 , then

HCF ( a,b) = HCF ( b,r)

Again expressing the integer b,r in Euclid’s Division Lemma, we get

b=pr + r_{1}

HCF ( b,r)=HCF ( r,r_{1})

Similarly successive Euclid ‘s division can be written until we get the remainder zero, the divisor at that point is called the HCF of the a and b

We will using the same in this questionFor a and b any two positive integer, we can always find unique integer q and r such that

a=bq + r , 0 ≤ r ≤ b

If r =0, then b is divisor of a.

HCF of two positive integers can be find using the Euclid’s Division Lemma algorithm

We know that for any two integers a,b. we can write following expression

a=bq + r , 0 ≤ r ≤ b

If r=0 ,then

HCF( a,b) =b

If r≠0 , then

HCF ( a,b) = HCF ( b,r)

Again expressing the integer b,r in Euclid’s Division Lemma, we get

b=pr + r

HCF ( b,r)=HCF ( r,r

Similarly successive Euclid ‘s division can be written until we get the remainder zero, the divisor at that point is called the HCF of the a and b

i) We should always start with Larger number

So in this case the larger number is 225

Now we can write 135 and 225 in Euclid division algorithm

225=135 X 1 +90

Now HCF of numbers (225,135) = HCF ( 135,90)

Again writing 135,90 is Euclid division formula

135=90X1 +45

Now HCF (135,90)= HCF (90,45)

Again writing 90,45 is Euclid division formula

90=45X2 +0

Now r =0, 45 is HCF (90,45)

45 is HCF ( 225,135)

ii) We should always start with Larger number

So in this case the larger number is 38220

Now we can write 38220 and 196 in Euclid division algorithm

38220=196 X 195 +0

Now r =0, 196 is HCF (196,38220)

iii) We should always start with Larger number

So in this case the larger number is 867

Now we can write 867 and 255 in Euclid division algorithm

867=255 X 3 +102

Now HCF of numbers (867,255) = HCF ( 255,102)

Again writing 255,102 is Euclid division formula

255=102X2 +51

Now HCF (255,102)= HCF (102,51)

Again writing 102,51 is Euclid division formula

102=51X2 +0

Now r =0, 51 is HCF (102,51)

51 is HCF ( 867,255)

For a and b any two positive integer, we can always find unique integer q and r such that

a=bq + r , 0 ≤ r < b

Now on putting b=6 ,we get

a=6q+ r , 0 ≤ r < 6

a=6q , This is a even number

a=6q+1 , This is a odd number

a=6q+2 ,This is an even number as 6 and 2 are divisible by 2

a=6q+3, it is not divisible by 2

a=6q+4, it is divisible by 2

a=6q+5 , it is not divisible by 2

So 6q,6q+2,6q +4 are even number

6q+1,6q+3,6q+5 are odd numbers

According to the questions, we need to find the maximum number of column

Maximum number of column is the HCF of number 32 and 616

So question has reduced to finding the HCF of 32 and 616 using Euclid division algorithm

We should always start with Larger number

So in this case the larger number is 616

Now we can write 616 and 32 in Euclid division algorithm

616=32 X19 +8

Now HCF of numbers (616,32) = HCF ( 32,8)

Again writing 32,8 is Euclid division formula

32=8X4+0

As r=0, 8 is the HCF of 616 and 32

For a and b any two positive integer, we can always find unique integer q and r such that

a=bq + r , 0 ≤ r < b

Now on putting b=3 ,we get

a=3q+ r , 0 ≤ r < 3

a=3q , a

a=3q+1 , a

a=3q+2 , a

For a and b any two positive integer, we can always find unique integer q and r such that

a=bq + r , 0 ≤ r < b

Now on putting b=3 ,we get

a=3q+ r , 0 ≤ r < 3

a=3q , a

a=3q+1 , a

a=3q+2 , a

Download Real Numbers Exercise 1.1 as pdf

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